December 8th
Today I learned the definition of generalized eigenvectors, to move towards Jordan normal form sometime in my far future. The idea is to patch the fact that sometimes we can't make a full eigenbasis. So, for example, the horizontal shear\[A=\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}\]has only a dimension-one eigenspace, spanned by $\langle1,0\rangle.$ Explicitly, its characteristic polynomial is $(\lambda-1)^2,$ so $1$ is the only eigenvalue. Its eigenspace is\[\ker(A-1I)=\ker\left(\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}\right),\]which evaluates to $\{\langle x,0\rangle:x\in\RR\},$ indeed spanned by $\langle1,0\rangle.$ While there are no more eigenvectors, we can get close to being an eigenvector by chaining, solving\[\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}v=\begin{bmatrix} 1 \\ 0 \end{bmatrix}.\]This gives $v=\langle x,1\rangle$ for any other $x.$ And further, our $\langle1,0\rangle$ and $\langle0,1\rangle$ (say) from our "chain'' does indeed form a basis of $\RR^2.$ So while we don't have an eigenbasis, we do have a generalized eigenbasis.
With this in mind, I should probably define what a generalized eigenvector is. We say that $v$ is a generalized eigenvector of rank $m$ if and only if $m$ is the smallest positive integer for which\[(A-\lambda I)^mv=0.\]Eigenvectors, then, are generalized eigenvectors of rank $1.$ In the above example, $\langle0,1\rangle$ is a generalized eigenvector of rank $2.$
It will turn out that the dimension of the generalized eigenspace associated with an eigenvalue does indeed match the algebraic multiplicity of the eigenvalue, something which I do not currently know a proof of. (I expect chaining as above to be able to give a full basis.) Then it's my understanding that changing to a basis of these chains will give the Jordan normal form, but I have not learned the details of this argument.